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\(73! - 72! = 73 \times 72! - 1 \times 72! = (73 - 1) \times 72! = 72 \times 72!\)
\(72! \times 72 = 2^3 \times 3^2 \times 72!\)

Highest Power of 6
Since \(6 = 2 \times 3\), we need the minimum power of 2 and 3 in \(72! \times 72\).

- Highest power of 2 in \(72!\):

Using Legendre’s formula:

\(\sum_{k=1}^{\infty} \left\lfloor \frac{72}{2^k} \right\rfloor = \left\lfloor \frac{72}{2} \right\rfloor + \left\lfloor \frac{72}{4} \right\rfloor + \left\lfloor \frac{72}{8} \right\rfloor + \left\lfloor \frac{72}{16} \right\rfloor + \left\lfloor \frac{72}{32} \right\rfloor + \left\lfloor \frac{72}{64} \right\rfloor\)

\(= 36 + 18 + 9 + 4 + 2 + 1 = 70\)

Including the extra \(2^3\) from 72:

\(70 + 3 = 73\)

- Highest power of 3 in \(72!\):

\(\sum_{k=1}^{\infty} \left\lfloor \frac{72}{3^k} \right\rfloor = \left\lfloor \frac{72}{3} \right\rfloor + \left\lfloor \frac{72}{9} \right\rfloor + \left\lfloor \frac{72}{27} \right\rfloor\)

\(= 24 + 8 + 2 = 34\)

Including the extra \(3^2\) from 72:

\(34 + 2 = 36\)


The highest power of 6 is determined by:

\(min(73, 36) = 36\)

Option D - 36
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Hello ,
73!−72! =72!(73-1)=72!*72
so 72! will have maximum power of 34 of 6
72 will also have max power of 2 as 72 can written as 6*6*2

so total maximum power will be 36
Hence D is correct answer
I hope it Helps

Thank you !!
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